$$\arctan{1} + \arctan{\frac{1}{2}} + \arctan{\frac{1}{3}} + \arctan{\frac{1}{4}} ...= ?$$

The infinite series for arctan is

$$\arctan{x} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ...$$

So I want to sum up the $\arctan{1\over n}$ where $n$ starts at $1$ and goes to infinity.

I originally thought the resulting series can be written this way:

$$(1 + \frac{1}{2} + \frac{1}{3} ...) - \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^3}{3} + \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^5}{5} - \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^7}{7} ...$$

But that is wrong. The right way to "insert" the series is:

$$1 + \frac{1}{2} + \frac{1}{3} ... - \frac{1}{3} - \frac{(\frac{1}{2})^3}{3} - \frac{(\frac{1}{3})^3}{3} ... + \frac{1}{5} + \frac{(\frac{1}{2})^5}{5} + \frac{(\frac{1}{3})^5}{5} ... ...$$

So it looks like a bunch of harmonic serieses manipulated. Harmonic series diverges, but I remember that doesn't necessarily mean a similar series diverges. I remember from Calculus 2 that, for example, $\lim_{n\to\infty}$ of $\sin(n)\over n$ converges to zero even though $\sin(n)$ does not converge.

So how do I figure out what this series converges to, if anything?


4 Answers 4


We have,

$$\lim_{n \to \infty} \frac{\arctan(\frac{1}{n})}{\frac{1}{n}}=1$$

This follows easily by first the change of variables $\frac{1}{n}=h$ then by Taylor series.

We also have that $a_n=\arctan (\frac{1}{n}) \geq 0$ for all $n \geq 1$. Hence by the limit comparison test your series $\sum_n a_n$ diverges with comparison to the harmonic series.


Based on this (On the arctangent inequality.) answer, we have:

$$\frac{\arctan x}{x} \geq 1/2$$

for $x \in (0,1]$.

So letting $x = \frac{1}{n}$, we have $\arctan \frac{1}{n} \geq \frac{1}{2n}$ for each $n\geq 1$, so by the comparison test, your series diverges


For $0 < x \leq 1$, the series

$$ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ... $$

is an alternating series with decreasing terms. This means the terms alternate between overshooting and undershooting the actual value of $\arctan(x)$. In particular,

$$ 0 < x \leq 1 \implies x - \frac{x^3}{3} < \arctan(x) < x $$

We can use this to get a good bound on the partial sums

$$ \sum_{k=1}^n \arctan\left( \frac{1}{k} \right) < \sum_{k=1}^n \frac{1}{k} = H_n$$

$$ \sum_{k=1}^n \arctan\left( \frac{1}{k} \right) > \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{3 k^3} \right) = H_n - \sum_{k=1}^n \left( \frac{1}{3 k^3} \right) > H_n - \frac{1}{3} \zeta(3)$$

So not only does the infinite sum go to infinity, it does so in basically the same fashion as harmonic numbers $H_n$ do, and furthermore the error in this estimate is strictly less than $0.4007$.


It seems it has not yet been pointed out that your rearrangement does not work!

$\frac11 - \frac12 + \frac13 - \frac14 + \cdots = \ln(2) > \frac12$.

$\frac11 - \frac12 + \frac13 - \frac14 + \cdots \ne \color{red}{( \frac11 + \frac13 + \frac15 + \frac17 + \cdots ) - ( \frac12 + \frac14 + \frac16 + \frac18 \cdots )}$.   [RHS is ill defined!]

$\frac11 - \frac12 + \frac13 - \frac14 + \cdots $

$\ \ne ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots ) + \frac13 ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots )$

$\ \quad + \frac1{3^2} ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots ) + \frac1{3^3} ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots ) + \cdots$

$\ = \frac12 \times ( \frac1{1 \times 2} - \frac1{4 \times 5} + \frac1{7 \times 8} - \cdots ) < \frac14$.

  • 1
    $\begingroup$ I seem to recall there's a theorem about what kinds of series converge regardless of re-ordering; you appear to be appealing to that theorem. $\endgroup$ Mar 7, 2017 at 19:16
  • 1
    $\begingroup$ @CarlWitthoft: It's just a matter of making sure the partial sums converge to the same value. In general absolute convergence is equivalent to convergence under any rearrangement. Conditional convergent series must be manipulated carefully to preserve limits. My example does use the right manipulations wherever I claim equality. =) $\endgroup$
    – user21820
    Mar 8, 2017 at 6:47
  • $\begingroup$ I don't know what "ill defined" means, but I can tell you I chose my arrangement just as a practical way to calculate the series. I wanted to calculate it out to 3, 4, 5, etc terms, just to see if I could find something it appeared to be converging too. I think when the terms are limited, it is defined with no ambiguity. $\endgroup$
    – DrZ214
    Mar 8, 2017 at 8:06
  • $\begingroup$ @DrZ214: No I've already stated precisely why your rearrangement is logically wrong. Whether or not you get the right answer with a wrong method does not make it correct. My post gives an explicit example where rearrangement of a convergent series in a wrong way can lead to a convergent series that converges to a different value. You are doing a similarly wrong manipulation that you called "the right way". You are not allowed to rearrange any infinite sum any way you like, even if the resulting terms are finite. Please revise the definition of the infinite sum. $\endgroup$
    – user21820
    Mar 8, 2017 at 13:45
  • $\begingroup$ @DrZ214: It is the limit of the partial sums if that limit exists. If you rearrange a sequence, you are changing the sequence of partial sums, so you may change the limiting behaviour. Not only can it change from "limit exists" to "limit does not exist" or vice versa, it may change from "limit equals 1" to "limit equals 2"! So no go; you must be completely sure of what you are doing when dealing with infinite series! $\endgroup$
    – user21820
    Mar 8, 2017 at 13:48

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